- #1
mugzieee
- 77
- 0
A 300-m tall tower is built on the equator. How much faster does a point at the top of the tower move than a point at the bottom?
Velocity in uniform circular motion
- Thread starter mugzieee
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Homework Help Overview
The discussion revolves around a problem in circular motion, specifically calculating the difference in linear velocity between two points on a tower located at the equator, one at the top and one at the bottom. The context involves understanding the effects of Earth's rotation on these velocities.
Discussion Character
- Exploratory, Assumption checking
Approaches and Questions Raised
- Participants discuss the calculation of velocities based on the height of the tower and Earth's rotation period. Questions arise regarding the period of Earth's rotation and its relevance to the problem.
Discussion Status
The discussion includes attempts to clarify the period of Earth's rotation, with one participant confirming the correctness of a calculated answer. There is an acknowledgment of a question that was later deemed unnecessary.
Contextual Notes
Participants are working within the constraints of a homework problem, which may limit the depth of exploration regarding the physics concepts involved.
- #2
xanthym
Science Advisor
- 410
- 0
{Period of Earth's Rotation} = T = (86400 sec)mugzieee said:
{Radius of Earth} = R
{Tower Bottom Speed} = 2*Pi*R/T
{Tower Top Speed} = 2*Pi*(R + 300 meters)/T
{Tower Top Speed} - {Tower Bottom Speed} = {2*Pi*(R + 300 meters)/T} - {2*Pi*R/T} =
= 2*Pi*(300)/T
= 2*Pi*(300)/(86400)
= (0.02182 m/sec)
~~
- #3
mugzieee
- 77
- 0
hey bro how did you get the T=86400 sec? thanks for your help, it was the correct answer!
- #4
brewnog
Science Advisor
Gold Member
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- 8
Mugzieee, how long is a day?
- #5
mugzieee
- 77
- 0
sorry, that was a thoughtless question which i realized afterwards. thank you 
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